What changes when an integral’s boundary is x squared?
You will be able to: Combine FTC with the chain rule, including reversed and two moving bounds.
What changes when an integral’s boundary is x squared?
Imagine a recording that plays twice as fast: the recorded process and playback speed both affect how quickly the displayed total changes. A moving integral boundary also contributes its own rate.
A useful starting point: Why does the derivative of accumulated change recover the rate? →
Words and symbols before equations
- g(x)
- A moving integration boundary.
- Chain rule
- The derivative of a composition multiplies by the inner derivative.
- Upper boundary
- The endpoint contributing positively to an integral.
- Lower boundary
- The starting endpoint, contributing with a minus sign when it moves.
What this picture assumes
Original model; readouts are rounded. B(x)=∫₀ˣ²(1+t)dt=x²+x⁴/2. B′(x)=2x(1+x²). Upper boundary x² moves in different directions for negative and positive x.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x=1; upper bound x²=1; B=1.5; B′=(1+x²)2x=4. Boundary speed 2x=2.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
If B(x)=∫₀ᵍ⁽ˣ⁾ f(t)dt, view it as A(g(x)). FTC gives A′=f, then the chain rule gives B′(x)=f(g(x))g′(x). The boundary speed g′ cannot be omitted.
For B(x)=∫₀ˣ² (1+t)dt, B′=(1+x²)2x. At x=1, this is 4; at x=−1 it is −4 even though the upper-bound integrand value is positive.
Reversing bounds changes the integral’s sign. Therefore the derivative of the integral from g(x) to a of f(t)dt=−f(g(x))g′(x).
For two moving bounds, split at a fixed reference point: the derivative of the integral from u(x) to v(x) of f(t)dt=f(v(x))v′(x)−f(u(x))u′(x). Assume continuous f on the needed range and differentiable boundaries.
A worked example, step by step
Differentiate H(x)=∫ₓˣ² (t+1)dt.
- Both boundaries move, so account for both contributions.
- Upper contribution: (x²+1)×2x.
- Lower contribution: subtract (x+1)×1.
- H′(x)=2x(x²+1)−(x+1); at x=2 the value is 20−3=17.
Replacing t by the bound is only half the chain rule. Multiply by the boundary derivative and subtract the lower-bound contribution.
If g′(x)=0, can the derivative of the accumulated composition be zero even when f(g(x)) is positive?
Compare with an explanation
Yes. A stationary boundary gives zero instantaneous change through the chain rule.
Predict. Change one thing. Explain.
Move x from −2 to 2 for B(x)=∫₀ˣ²(1+t)dt. Explain why B is even but B′ changes sign. Compare the upper-bound position with its direction of motion.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=1; upper bound x²=1; B=1.5; B′=(1+x²)2x=4. Boundary speed 2x=2.
Original model; readouts are rounded. B(x)=∫₀ˣ²(1+t)dt=x²+x⁴/2. B′(x)=2x(1+x²). Upper boundary x² moves in different directions for negative and positive x.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, signed contributions, theorem conditions, domain or antiderivative check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionDifferentiate J(x), the integral from x² to 2x of (1+t²)dt, and evaluate J′(1).
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Compare with the answer and four-point rubric
- 1 point: Continuous integrand and differentiable boundaries permit FTC and chain rule.
- 1 point: Upper contribution is 2(1+4x²).
- 1 point: Subtract lower contribution 2x(1+x⁴).
- 1 point: At x=1, J′=10−4=6.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which factor is often missed?
The derivative of the moving boundary.
RECALL 2Why does the lower bound have a minus sign?
Moving the starting boundary removes that oriented contribution.
RECALL 3What if both boundaries move?
Compute upper contribution minus lower contribution.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What changes when an integral’s boundary is x squared?
- d/dx ∫ₐᵍ⁽ˣ⁾ f(t)dt=f(g(x))g′(x).
- Two moving bounds: upper contribution minus lower contribution.
Remember: Replacing t by the bound is only half the chain rule. Multiply by the boundary derivative and subtract the lower-bound contribution.
Conditions: Original model; readouts are rounded. B(x)=∫₀ˣ²(1+t)dt=x²+x⁴/2. B′(x)=2x(1+x²). Upper boundary x² moves in different directions for negative and positive x.
Refresh Kid · AP Calculus AB Unit 6 · Objectives FUN-5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.4, FUN-5.A. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. AB scope includes 6.1–6.10 and 6.14. Topics 6.11–6.13 are BC-only; the numbering gap is intentional. Topic 6.14 consolidates the AB antidifferentiation objectives and Skill 1.C. Focused lesson titles, examples and questions are original Refresh Kid material.
Distinguish signed accumulation, unsigned area and initial amount. Error direction needs interval-wide shape information. FTC hypotheses, variable-bound chain factors, integration constants, transformed bounds and domain restrictions are explicit. Bounded jumps are treated separately from differentiability of accumulation. No improper-integral evaluation, integration by parts or partial-fraction decomposition is taught in this AB unit.
The Organic Chemistry Tutor Fundamental Theorem of Calculus Part 1 video title, creator and description were checked; the full video was not reviewed. Khan Academy’s unit destination was checked, but its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 5.2, 5.3 and 5.5 were consulted for conceptual cross-checking. No provider scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank geometry uses existing self-hosted Three.js with its MIT license. Its 2×2 dm base and water depth use the same volume as the rate calculation: 1 dm³=1 L. The initial water is blue and newly accumulated inflow is teal. The rate graph is an abstract amount calculation; it is not the physical shape of water. Camera rotation changes only the view. Full 2D graphs, readouts and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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